Question

1 g molecule of N2 and 4 g molecules of H2 are mixed in a 2 L vessel and are allowed to prepare NH3(g) through Haber’s process. Which of the following acidic solution will exactly neutralise this NH3(g) solution?

Answer 1

Given:

1 g molecule of N2 and 4 g molecules of H2 are mixed in a 2 L vessel and are allowed to prepare NH3(g) through Haber’s process.

To find:

Concentration of acid that could exactly neutralize the ammonia solution.

Calculation:

$\boxed{\sf{N_{2}+3H_{2}\rightarrow 2NH_{3}}}$

Moles of N2 = 1/28 moles

Moles of H2 = 4/2 = 2 moles

Here nitrogen is the limiting reagent;

1 mole of N2 gives 2 moles of NH3

=> 1/28 moles of N2 gives = 2/28 moles of NH3

=> 1/28 moles of N2 gives 1/14 moles of NH3.

So, concentration of NH3:

$molarity = \dfrac{( \frac{1}{14}) }{2}$

$= > molarity = \dfrac{1}{28} \: M$

So, any acid releasing this concentration of protons will exactly neutralize the ammonia produced through Habers process.

So, final answer is:

$\boxed{ \sf{ conc. \: of \: {H}^{ + } = \dfrac{1}{28} \: M }}$